\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 80, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/80\hfil Positive solutions]
{Positive solutions for fractional differential equations
 with variable coefficients}

\author[Y. Chen\hfil EJDE-2012/80\hfilneg]
{Yi Chen, Zhanmei Lv}

\address{Yi Chen \newline
School of Mathematical Sciences and Computing Technology\\
Central South University\\
Changsha, Hunan 410083, China}
\email{mathcyt@163.com}

\address{Zhanmei Lv \newline
School of Mathematical Sciences and Computing Technology,
Central South University, 
Changsha, Hunan 410083, China}
\email{cy2008csu@163.com}

\thanks{Submitted April 3, 2012. Published May 18, 2012.}
\thanks{Supported by grant CX2011B079 from the Hunan Provincial Innovation
 Foundation \hfill\break\indent for Postgraduate.}
\subjclass[2000]{26A33, 34A08, 34A12}
\keywords{Fractional differential equations;  fixed point theorem; 
\hfill\break\indent
positive solution; multiplicity solution}

\begin{abstract}
 In this article, we study the existence of the positive solutions
 for a class of differential equations of fractional order with variable
 coefficients. The equation of this type plays an important role in the
 description and modeling of control systems, such as $PD^{\mu}$-controller.
 The differential operator is taken in the Riemann-Liouville sense.
 Our analysis relies on the Leggett-Williams fixed point theorem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

Fractional calculus is a generalization of the ordinary differentiation and 
integration. It plays an important role in science, engineering, economy, 
and other fields, see \cite{d1,h1, k1,l1,m1,m2,o1,p1}.
 For example, the book \cite{m2} details the use of fractional calculus 
in the description and modeling of systems, and in a range of control design 
and practical applications. And today there are many papers dealing with the 
fractional differential equations due to its various applications, 
see \cite{a1,a2,b1,b2,b3,l3,n1,s1,s2,s3,z1,z2}.

In \cite{m2}, the authors considered the dynamic model of an immersed plate,
 which is modeled by
\begin{equation} \label{e1}
\begin{gathered}
A_BD^{2}_{0+}y(t)+B_BD^{1.5}_{0+}y(t)+C_By(t)=f(t),\nonumber\\
y(0)=y'(0)=0.
\end{gathered}
\end{equation}
As indicated in \cite{p1}, a fractional order $PD^{\mu}$-controller can
 be more suitable for the control of "reality" than integer order.
For example, the fractional-order $PD^{\mu}$-controller can be characterized
 by (see \cite[equation (9.33)]{p1})
\begin{equation}\label{e2}
a_2D^{\beta}_{0+}y(t)+T_dD^{\mu}_{0+}y(t)+a_1D^{\alpha}_{0+}y(t)+(a_0+K)y(t)
=Kw(t)+T_dD^{\mu}_{0+}w(t),
\end{equation}
where $\alpha<\mu<\beta$. And, \eqref{e1} and \eqref{e2} are the particular
case of the equation of type \eqref{maineq} in our paper. And for the system
of this type, we can find its many other real applications in
\cite{o1,p1,r1} and in \cite[Chapter 14-18]{m2}.

Problems of this type, with constant coefficients, have provoked some interest 
in recent literature, such as \cite{b1,s1,s2,s3} and references therein. 
In \cite{s1}, the author indicated that: 
``Some of the earlier results of this type contains errors in the proof 
of equivalence of the initial value problems and the corresponding Volterra 
integral equations (see survey paper by Kilbas and Trujillo \cite{k2})".

Motivated by these papers, in this paper, we  consider the following initial 
value problems of fractional differential equations with variable coefficients
\begin{equation}\label{maineq}
\begin{gathered}
 D^{\alpha_n}_{0+}u(t)-\sum_{j=1}^{n-1}a_{j}(t)D^{\alpha_{j}}_{0+}u(t)
 =f(t,u(t)),\quad 0 \leq t \leq 1,\\
 u(0)=u'(0)=0,
\end{gathered}
\end{equation}
 where $0< \alpha_1 < \alpha_2 < \dots < \alpha_{n-1}< \alpha_n-1< 1 < \alpha_{n}
 < 2$,
$n \geq 2$, $n \in \mathbb{Z}$, $a_n \in \mathbb{R}$,
$f:[0,1]\times [0,+\infty)\to [0,+\infty)$ is continuous and 
$a_j:[0,1]\to (0,+\infty)$ ($j=1,2,\dots,n-1$) are continuously differentiable. 
We will study the problem \eqref{maineq} in the Banach space $C[0,1]$ equipped
 with the maximum norm $\|\cdot\|$.

To the best of our knowledge, the results on the existence of solutions 
for the fractional differential equations with variable coefficients are 
relatively scare. The variable coefficients cause the problem more complex. 
The main difficulty in dealing with such issues is that the classical 
integration by parts formula is no longer applicable for the fractional integration. 
And how to get the equivalent integral equation of the problem \eqref{maineq} 
differs from the equations with constant coefficients. In the paper we solve 
these problems.

This article is organized as follows. In Section 2, we present some results 
of fractional calculus theory and auxiliary technical lemmas, which are used 
in the next section. Section 3, applying the results of Section 2, we obtain 
the existence and multiplicity results of the positive solutions for the 
problem \eqref{maineq} by the Leggett-Williams fixed point theorem in a cone. 
Then an example is given in Section 4 to demonstrate the application 
of our results.


\section{Preliminaries}

First of all, we present the necessary definitions and fundamental facts 
on the fractional calculus theory. These can be found in \cite{k1,m1,p1}.

\begin{definition}[\cite{k1,o1,p1}] \label{def2.1} \rm
The Riemann-Liouville fractional integral of order $\nu>0$ of a function 
$h:(0, \infty)\to \mathbb{R}$ is given by
\begin{equation}
I^{\nu}_{0+}h(t)=D^{-\nu}_{0+}h(t)=\frac{1}{\Gamma(\nu)}\int^t_0(t-s)^{\nu-1}h(s)ds
\end{equation}
provided that the right-hand side is pointwise defined on $(0, \infty)$.
\end{definition}


\begin{definition}[\cite{k1,o1,p1}]\label{def2.2} \rm
The Riemann-Liouville fractional derivative of order $\nu>0$ of a continuous 
function $h:(0, \infty)\to \mathbb{R}$ is given by
\begin{equation}
D^{\nu}_{0+}h(t)=\frac{1}{\Gamma(n-\nu)}\big(\frac{d}{dt}\big)^n
\int^t_0(t-s)^{n-\nu-1}h(s)ds,
\end{equation}
where $n=[\nu]+1$, provided that the right-hand side is pointwise defined on
 $(0, \infty)$.
\end{definition}

\begin{lemma}[\cite{b3}]\label{lem2.1}
Assume that $h(t) \in C(0, 1)\cap L(0, 1)$ with a fractional derivative
 of order $\nu>0$ that belongs to $C(0, 1)\cap L(0, 1)$. Then
\begin{equation}
I^{\nu}_{0+}D^{\nu}_{0+}h(t)=h(t)+C_1t^{\nu-1}+C_2t^{\nu-2}+ \dots +C_Nt^{\nu-N},
\end{equation}
for some $C_i \in \mathbb{R}$, $i=1,2,\ldots,N$, where $N$ is the smallest 
integer such that $N \geq \nu$.
\end{lemma}

\begin{lemma}[\cite{k1,o1,p1}]\label{lem2.2}
 If $\nu_1,\nu_2,\nu>0$, $t \in [0,1]$ and $h(t) \in L[0,1]$, then 
\begin{equation}
I^{\nu_1}_{0+}I^{\nu_2}_{0+}h(t)=I^{\nu_1+\nu_2}_{0+}h(t),\quad
D^{\nu}_{0+}I^{\nu}_{0+}h(t)=h(t).
\end{equation}
\end{lemma}

\begin{lemma}[\cite{l3,p1}]\label{lem2.3}
 If $h(t) \in C[0,1]$ and $\nu>0$, then we have
\begin{equation}
\left[I^{\nu}_{0+}h(t) \right]_{t=0}=0,\quad\text{or}\quad 
\lim_{t\to 0}\frac{1}{\Gamma(\nu)}\int^t_0(t-s)^{\nu-1}h(s)ds=0.
\end{equation}
\end{lemma}

Let
\begin{gather}\label{gi}
g_j(t,s)=(\alpha_n-1)a_j(s)-(t-s)a_j'(s),\quad (t,s) \in [0,1]\times[0,1],\\
\label{hj}
h_j(t,\tau)=\int_0^1{\xi}^{-\alpha_{j}}(1-\xi)^{\alpha_{n}-2}g_j(t,\tau+\xi(t-\tau))
d\xi,\quad (t,\tau) \in [0,1]\times[0,1],
\end{gather}
where $j=1,2,\dots,n-1$.
It is obvious that $g_j(t,s),h_j(t,\tau)$ are continuous and that for $0<s<t$,
\begin{align*}
\frac{d}{ds}\big((t-s)^{\alpha_n-1}a_j(s)\big)
&= -(\alpha_n-1)(t-s)^{\alpha_n-2}a_j(s)+(t-s)^{\alpha_n-1}a_j'(s)\\
&= -(t-s)^{\alpha_n-2}g_j(t,s),\quad j=1,2,\dots,n-1.
\end{align*}
Set
\begin{equation}\label{bj}
b_j(t)=\ln{a_j(t)},\quad j=1,2,\dots,n-1,
\end{equation}
then it is clear that $b_j(t)$ $(j=1,2,\dots,n-1$) is continuously differentiable.

\begin{lemma}\label{lem2.4}
Let $a_j:[0,1]\to (0,+\infty)$ $(j=1,2,\dots,n-1)$ are continuously differentiable. 
Assume that the condition
\begin{itemize}
\item[(H1)] $|{b_j}'(t)|< \alpha_n-1$, $j=1,2,\dots,n-1$.
\end{itemize}
 Then  $g_j(t,s)>0$, for $j=1,2,\dots,n-1$.
\end{lemma}

\begin{proof}
In view of \eqref{bj}, we have
\begin{equation*}
a_j(t)=e^{b_j(t)},\ \ {a_j}'(t)={b_j}'(t)e^{b_j(t)},\ \ j=1,2,\dots,n-1.
\end{equation*}
\par
Then, by $(H_1)$, we deduce that
\begin{align*}
g_j(t,s)&= (\alpha_n-1)a_j(s)-(t-s)a_j'(s)\\
&= (\alpha_n-1)e^{b_j(t)}-(t-s)b_j'(t)e^{b_j(t)}\\
&=e^{b_j(t)} \big((\alpha_n-1)-(t-s)b_j'(t)\big)>0.
\end{align*}
The proof is complete.
\end{proof}

For convenience, denote
\begin{gather*}
M_j=\max_{0\leq t \leq 1,\, 0\leq s \leq 1}g_j(t,s),\quad
m_j=\min_{0\leq t \leq 1,\, 0\leq s \leq 1}g_j(t,s),\quad j=1,2,\dots,n-1;
\\
P_1=\sum_{j=1}^{n-1}\frac{M_j B(1-\alpha_j,\alpha_n-1)}{(\alpha_n-\alpha_j)
\Gamma{(\alpha_n)}\Gamma{(1-\alpha_j)}}
=\sum_{j=1}^{n-1}\frac{M_j}{(\alpha_n-1)\Gamma{(\alpha_n-\alpha_j+1)}};
\\
P_2=\sum_{j=1}^{n-1}\frac{m_j B(1-\alpha_j,\alpha_n-1)}{(\alpha_n-\alpha_j)
\Gamma{(\alpha_n)}\Gamma{(1-\alpha_j)}}
=\sum_{j=1}^{n-1}\frac{m_j}{(\alpha_n-1)\Gamma{(\alpha_n-\alpha_j+1)}};
\end{gather*}
We can  easily show that $M_j \geq m_j >0$ and $P_1\geq P_2 >0$. 
Then we have the following lemma.

\begin{lemma}\label{lem2.5}
Let $f:[0,1]\times [0,+\infty)\to [0,+\infty)$ is continuous and
 $a_j:[0,1]\to (0,+\infty)\ (j=1,2,\dots,n-1)$ are continuously differentiable,
 then 
\begin{equation}\label{est_h}
m_j B(1-\alpha_j,\alpha_n-1)\leq h_j(t,\tau)\leq M_j B(1-\alpha_j,\alpha_n-1),\quad
 j=1,2,\dots,n-1,
\end{equation}
where $B(\cdot,\cdot)$ is the Beta function.
\end{lemma}

\begin{proof}
According to \eqref{hj}, for each $j=1,2,\dots,n-1$, we have
\[
h_j(t,\tau)
 \leq  M_j \int_0^1{\xi}^{-\alpha_{j}}(1-\xi)^{\alpha_{n}-2}d\xi
 = M_j B(1-\alpha_j,\alpha_n-1),\quad (t,\tau) \in [0,1]\times[0,1].
\]
Analogously,
\begin{align*}
 h_j(t,\tau) \geq m_j B(1-\alpha_j,\alpha_n-1),\ (t,\tau) \in [0,1]\times[0,1].
\end{align*}
Then we obtain \eqref{est_h}.
\end{proof}

\begin{definition} \rm
$u(t) \in C[0,1]$ is called a solution of the problem \eqref{maineq} 
if $u'(t)$ exists in $[0,1]$ and $u(t)$ satisfied the equation and the initial 
conditions in \eqref{maineq}.
\end{definition}

\begin{lemma}\label{lem1}
Let $f:[0,1]\times [0,+\infty)\to [0,+\infty)$ is continuous and
 $a_j:[0,1]\to (0,+\infty)$ $(j=1,2,\dots,n-1)$ are continuously differentiable, 
then $u(t)$ is a solution of the equation \eqref{maineq} if and only if
 $u(t)\in C[0,1]$ is the solution of the integral equation
\begin{equation}\label{equivlentequation}
\begin{split}
u(t)&=\sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}
\int_0^t(t-\tau)^{\alpha_{n}-\alpha_{j}-1}u(\tau)h_j(t,\tau) d\tau\\
&\quad + \frac{1}{\Gamma({\alpha_{n}})}
\int_0^t(t-\tau)^{\alpha_{n}-1}f(\tau,u(\tau))d\tau
\end{split}
\end{equation}
\end{lemma}

\begin{proof}
``Necessity''. Applying Lemma \ref{lem2.1} and the initial conditions, we have
\begin{equation}\label{prooflem1}
u(t)= \lambda_1\sum_{j=1}^{n-1}I_{0+}^{\alpha_n}
\left(a_j(t)D_{0+}^{\alpha_j}u(t)\right)+\lambda_2I_{0+}^{\alpha_n}f(t,u(t)).
\end{equation}
Combining Definition \ref{def2.2} and Lemma \ref{lem2.3}, we have
\begin{align*}
&I_{0+}^{\alpha_n}\big(a_j(t)D_{0+}^{\alpha_j}u(t)\big)\\
&=\frac{1}{\Gamma{(\alpha_n)}}\int_0^t(t-s)^{\alpha_n-1}a_j(s)
\Big(\frac{d}{ds}\frac{1}{\Gamma{(1-\alpha_j)}}\int_0^
s(s-\tau)^{-\alpha_j}u(\tau)d\tau\Big)ds\\
& =\frac{1}{\Gamma{(\alpha_n)}\Gamma{(1-\alpha_j)}}
 \Big((t-s)^{\alpha_n-1}a_j(s)\int_0^s(s-\tau)^{-\alpha_j}u(\tau)d\tau\Big)
\Big|_{s=0}^{s=t}\\
&\quad   -\frac{1}{\Gamma{(\alpha_n)}\Gamma{(1-\alpha_j)}}
 \int_0^t\int_0^s\frac{d}{ds}\Big((t-s)^{\alpha_n-1}a_j(s)\Big)
 (s-\tau)^{-\alpha_j}u(\tau)d\tau ds\\
& =\frac{1}{\Gamma{(\alpha_n)}\Gamma{(1-\alpha_j)}}
  \int_0^t\int_0^s(t-s)^{\alpha_n-2}g_j(t,s)(s-\tau)^{-\alpha_j}u(\tau)d\tau ds\\
& =\frac{1}{\Gamma{(\alpha_n)}\Gamma{(1-\alpha_j)}}
 \int_0^t\int_{\tau}^t(t-s)^{\alpha_n-2}g_j(t,s)(s-\tau)^{-\alpha_j}u(\tau) ds\,d\tau\\
&=\frac{1}{\Gamma{(\alpha_n)}\Gamma{(1-\alpha_j)}}
 \int_0^t(t-\tau)^{\alpha_{n}-\alpha_{j}-1}u(\tau)
 \int_0^1{\xi}^{-\alpha_{j}}(1-\xi)^{\alpha_{n}-2}\\
&\quad\times g_j(t,\tau+\xi(t-\tau))d\xi d\tau\\
&=\frac{1}{\Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}
 \int_0^t(t-\tau)^{\alpha_{n}-\alpha_{j}-1}u(\tau)h_j(t,\tau)d\tau,
\end{align*}
where $j=1,2,\dots,n-1$. 
Thus, in view of  \eqref{prooflem1}, we derive \eqref{equivlentequation}.

``Sufficiency''. Suppose that $u(t)\in C[0,1]$ is the solution of 
\eqref{equivlentequation}. Then we canshow that $u'(t)$ exists in 
$[0,1]$ by \eqref{equivlentequation}. Also by exploiting Lemma \ref{lem2.2} and
 Lemma \ref{lem2.3}, we can deduce the equation \eqref{maineq} easily and prove 
that the initial conditions are satisfied. This completes the proof.
\end{proof}

\begin{theorem}[\cite{l2}]\label{thm3}
Let $(E,\|\cdot\|)$ be a Banach space, $P\subset E$ be a cone of $E$ and
 $c>0$ be a constant. Suppose that there exists a concave nonnegative continuous
 functional $\omega$ on $P$ with $\omega(x) \leq \|x\|$ for all 
$x \in \overline{P}_c$. Let $B: \overline{P}_c \to \overline{P}_c$
 be a completely continuous operator. Assume there are numbers $a, b$
 and $d$ with $0<d<a<b\leq c$ such that
\begin{itemize}
\item[(1)] $\{x\in P(\omega,a,b):\omega(x)>a\} \neq \emptyset$ and
  $\omega(Bx)>a$ for all $x\in P(\omega,a,b)$;
\item[(2)] $\|Bx\|<d$ for all $x\in \overline{P}_d$;
\item[(3)] $\omega(Bx)>a$ for all $x\in P(\omega, a, c)$ with $\|Bx\|>b$.
\end{itemize}
Then $B$ has at least three fixed points $x_1,\ x_2$ and $\ x_3$ in
 $\overline{P}_c$. Furthermore, 
$x_1 \in P_d$, $\ x_2 \in \{x \in P(\omega,a,c): \omega(x)>a\};$
$x_3 \in \overline{P}_c\backslash (P(\omega,a,c)\cup\overline{P}_d)$.
\end{theorem}

\section{Existence and multiplicity of  positive solutions}

Let
$$
K=\big\{x \in C[0,1]: x(t)\geq0,\, t \in [0,1],
 \min_{t \in [0,l_1]}x(t)\geq L \|x\|\big\},
$$
where $0< l_1 <1$ and $0<L<1$. Evidently, $K$ is a cone of the Banach 
space $C[0,1]$. In the following we will assume that $d/a<L<1$ 
(the constant $d$ and $a$ are defined in Theorem \ref{thm4.1}).
Define $\omega:K\to[0,+\infty)$ by
\begin{equation*}
\omega(u)=\min_{t \in [l_1,\,l_2]}u(t),\quad 0<l_1<l_2\leq 1.
\end{equation*}
It is easy to check that $\omega(u)$ is a concave nonnegative continuous 
functional on $K$, and satisfies $\omega(u)\leq \|u\|$ for all $u \in K$.

Denote $C^+[0,1]=\big\{x \in C[0,1]: x(t)\geq0,\ t \in [0,1]\big\}$.
 Then let us define three operators $A,B,T:C^+[0,1]\to C^+[0,1]$ as follows
\begin{gather*}
(A u)(t)= \sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})
\Gamma({1-\alpha_{j}})}\int_0^t(t-\tau)^{\alpha_{n}-\alpha_{j}-1}
 u(\tau)h_j(t,\tau) d\tau, \\
(B v)(t)= \frac{1}{\Gamma({\alpha_{n}})}\int_0^t(t-\tau)^{\alpha_{n}-1}
f(\tau,v(\tau))d\tau,
\\
(T \varphi)(t) =  (A \varphi)(t)+(B \varphi)(t),
\end{gather*}
where $u, v, \varphi \in C^+[0,1]$.


\begin{lemma}\label{lem4.1}
The operator $A:C^+[0,1]\to C^+[0,1]$ is continuous and compact.
\end{lemma}

\begin{proof}
Obviously, $A$ is continuous. So we only need to prove that $A$ is compact.
Let $U \subset C^+[0,1]$ be bounded; i.e., there exists a positive constant 
$r$ such that $\|u\| \leq r,\ \forall u \in U$, for each $u\in U$, 
via Lemma \ref{lem2.5}, we have
\begin{align*}
|(Au)(t)|
&= \Big|\sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_0^t(t-\tau)^{\alpha_{n}
 -\alpha_{j}-1}u(\tau)h_j(t,\tau) d\tau\Big|\\
&\leq \sum_{j=1}^{n-1}\frac{M_j B(1-\alpha_j,\alpha_n-1)}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_0^t(t-\tau)^{\alpha_{n}-\alpha_{j}-1}u(\tau)d\tau\\
&\leq \sum_{j=1}^{n-1}\frac{r M_j B(1-\alpha_j,\alpha_n-1)}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\frac{t^{\alpha_n-\alpha_j}}{\alpha_n-\alpha_j}\\
&\leq \sum_{j=1}^{n-1}\frac{r M_j B(1-\alpha_j,\alpha_n-1)}{(\alpha_n-\alpha_j)
 \Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}=P_1r.
\end{align*}
Thus $\|Au\|\leq P_1r$.
Hence $A(U)$ is bounded.

Next, let
\begin{equation*}
\gamma_1=2P_1r,\quad
\gamma_2=\sum_{j=1}^{n-1}\frac{r}{(\alpha_n-\alpha_j)
\Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}.
\end{equation*}
Since $h_j(t,\tau)\ (j=1,2,\dots,n-1)$ is uniformly continuous on
 $[0,1]\times[0,1]$, $\forall \varepsilon>0$, there exists a
 $\delta_j>0\ (\delta_j<1)$ such that
\begin{equation}
|h_j(t_1,\tau_1)-h_j(t_2,\tau_2)|\leq \frac{\varepsilon}{3\gamma_2},
\end{equation}
for all $(t_1,\tau_1),(t_2,\tau_2)\in[0,1]\times[0,1]$ with 
$|t_1-t_2|\leq\delta_j$ and  $|\tau_1-\tau_2|\leq\delta_j$, $j=1,2,\dots,n-1$.

Next prove that $A(U)$ is equicontinuous.
For the given $\varepsilon>0$, there exists $\rho_j>0(1\leq j\leq n-1)$ such that 
$|{t_2}^{\alpha_n-\alpha_j}-{t_1}^{\alpha_n-\alpha_j}|<\varepsilon/(3\gamma_1)$, 
where $|t_2-t_1|<\rho_j$. Let
$$
\delta=\min\big\{\delta_1,\ \delta_2,\ \dots,\ \delta_{n-1},\ \rho_1,\ \rho_{2},
\dots, \rho_{n-1},(\frac{\varepsilon}{3})^{1/{(\alpha_n-\alpha_{n-1})}}\big\}.
$$
For each $u \in U$, $t_1,t_2 \in [0,1]$ with $|t_1-t_2|\leq \delta  (t_1<t_2)$, 
we have
\begin{align*}
&|(Au)(t_1)-(Au)(t_2)|\\
&=\Big|\sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_0^{t_1}(t_1-\tau)^{\alpha_{n}
 -\alpha_{j}-1}u(\tau)h_j(t_1,\tau) d\tau\\
&\quad -\sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_0^{t_2}(t_2-\tau)^{\alpha_{n}
 -\alpha_{j}-1}u(\tau)h_j(t_2,\tau) d\tau\Big|\\ 
&\leq \Big|\sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_0^{t_1}
 [(t_1-\tau)^{\alpha_{n}-\alpha_{j}-1}-(t_2-\tau)^{\alpha_{n}-\alpha_{j}-1}]
 u(\tau)h_j(t_1,\tau) d\tau\Big|\\
&\quad +\Big|\sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}
 \int_0^{t_1}(t_2-\tau)^{\alpha_{n}-\alpha_{j}-1}u(\tau)[h_j(t_1,\tau)
 -h_j(t_2,\tau)] d\tau\Big|\\
&\quad +\Big|\sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_{t_1}^{t_2}(t_2-\tau)^{\alpha_{n}
 -\alpha_{j}-1}u(\tau)h_j(t_2,\tau) d\tau\Big|\\
& \leq\sum_{j=1}^{n-1}\frac{r M_j B(1-\alpha_j,\alpha_n-1)}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_0^{t_1}
 [(t_2-\tau)^{\alpha_{n}-\alpha_{j}-1}-(t_1-\tau)^{\alpha_{n}-\alpha_{j}-1}]d\tau\\
&\quad +\big(\frac{\varepsilon}{3\gamma_2}\big)
 \sum_{j=1}^{n-1}\frac{r}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_0^{t_1}(t_2-\tau)^{\alpha_{n}-\alpha_{j}-1}d\tau \\
&\quad +\sum_{j=1}^{n-1}\frac{r M_j B(1-\alpha_j,\alpha_n-1)}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_{t_1}^{t_2}(t_2-\tau)^{\alpha_{n}-\alpha_{j}-1}d\tau\\
&=\sum_{j=1}^{n-1}\frac{r M_j B(1-\alpha_j,\alpha_n-1)}{(\alpha_n-\alpha_j)
 \Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}[t_2^{\alpha_{n}-\alpha_{j}}
 -t_1^{\alpha_{n}-\alpha_{j}}-(t_2-t_1)^{\alpha_{n}-\alpha_{j}}]\\
&\quad +\big(\frac{\varepsilon}{3\gamma_2}\big)
 \sum_{j=1}^{n-1}\frac{r}{(\alpha_n-\alpha_j)
 \Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}
 [t_2^{\alpha_{n}-\alpha_{j}}-(t_2-t_1)^{\alpha_{n}-\alpha_{j}}] \\
&\quad+\sum_{j=1}^{n-1}\frac{r M_j B(1-\alpha_j,\alpha_n-1)}{(\alpha_n-\alpha_j)
 \Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}(t_2-t_1)^{\alpha_{n}-\alpha_{j}}\\
&\leq r \sum_{j=1}^{n-1}\frac{ M_j B(1-\alpha_j,\alpha_n-1)}{(\alpha_n-\alpha_j)
 \Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}
 [t_2^{\alpha_{n}-\alpha_{j}}-t_1^{\alpha_{n}-\alpha_{j}}]
 +\big(\frac{\varepsilon}{3\gamma_2}\big)\gamma_2\\
&\quad +r \sum_{j=1}^{n-1}\frac{M_j B(1-\alpha_j,\alpha_n-1)}{(\alpha_n-\alpha_j)
 \Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}(t_2-t_1)^{\alpha_{n}-\alpha_{j}}\\
&\leq r P_1 \big(\frac{\varepsilon}{3\gamma_1}\big)
+(r P_1)\big(\frac{\varepsilon}{3\gamma_1}\big)
 +\frac{\varepsilon}{3}\leq \frac{\varepsilon}{6}+\frac{\varepsilon}{6}
+\frac{\varepsilon}{3}< \varepsilon.
\end{align*}
Therefore, $A(U)$ is equicontinuous. And the Arzela-Ascoli theorem implies that 
$A(U)$ is relatively compact. Thus, the operator $A:C^+[0,1]\to C^+[0,1]$ is compact.
\end{proof}

\begin{lemma}\label{lem4.2}
The operator $B:C^+[0,1]\to C^+[0,1]$ is  continuous and compact.
\end{lemma}

\begin{proof}
It is obvious that the operator $B:C^+[0,1]\to C^+[0,1]$ is continuous.
 Similar to the proof of Lemma \ref{lem4.1}, by the Arzela-Ascoli theorem, 
we can conclude that the operator $B:C^+[0,1]\to C^+[0,1]$ is compact.
 Here we omit the proof.
\end{proof}

\begin{lemma}\label{lem4.3}
The operator $T:C^+[0,1]\to C^+[0,1]$ is continuous and compact.
\end{lemma}

The above lemma is obtained from Lemmas \ref{lem4.1} and  \ref{lem4.2}.
Now we present the main result of this article.


\begin{theorem}\label{thm4.1}
Let $f:[0,1]\times [0,+\infty)\to [0,+\infty)$ is continuous and 
$a_j:[0,1]\to (0,+\infty)\ (j=1,2,\dots,n-1)$ are continuously differentiable. 
Assume that $(H_1)$ holds and there exist three positive constants $0<d<a<b$ 
such that the following conditions are satisfied.
\begin{itemize}
\item[(H2)] $P_1<1$ and $f(t,u)\leq \Gamma{(\alpha_n+1)}(1-P_1)b$, for all 
$(t,u)\in[0,1]\times[0,b]$;

\item[(H3)] $C_1P_2L<1$,
and $f(t,u)\geq C_2\,a$, for all $(t,u)\in[0,l_1]\times[La,b]$, where
\begin{gather*}
C_1=\min\big\{\min_{t \in [l_1,\,l_2]}
[t^{\alpha_n-\alpha_j}-(t-l_1)^{\alpha_n-\alpha_j}],\ j=1,2,\dots,n-1\big\},\\
C_2> \frac{\Gamma{(\alpha_n+1)}(1-C_1P_2L)}{{l_1^{\alpha_n}}};
\end{gather*}

\item[(H4)] $P_1<1$ and $f(t,u)< \Gamma{(\alpha_n+1)}(1-P_1)d$, 
for all $(t,u)\in[0,1]\times[0,d]$.
\end{itemize}
Then problem \eqref{maineq} has at least three positive solutions
 $u_1, u_2$ and $ u_3$ in $\overline{K}_b$. Furthermore,
 $u_1 \in K_d$; $u_2 \in \{u \in K(\omega,a,b): \omega(u)>a\}$;
$u_3 \in \overline{K}_b\backslash (K(\omega,a,b)\cup\overline{K}_d)$.
\end{theorem}

\begin{proof}
From Section 2, we know that 
$K_d=\{u \in K: \|u\|<d\}$,
$\overline{K}_d=\{u \in K: \|u\|\leq d\}$,
 $\overline{K}_b=\{u \in K: \|u\|\leq b\}$ and 
$K(\omega,a,b)=\{u \in K: \omega(u) \geq a, \|u\|\leq b\}$.
We prove the results by three steps.

Step 1: $T: \overline{K}_b \to \overline{K}_b$ is a completely continuous operator.
 For any $u \in \overline{K}_b$, from $(H_2)$ and Lemma \ref{lem2.5}, we have
\begin{align*}
(T u)(t)
&=\sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}
 \int_0^t(t-\tau)^{\alpha_{n}-\alpha_{j}-1}u(\tau)h_j(t,\tau)d\tau\\
&\quad + \frac{1}{\Gamma({\alpha_{n}})}\int_0^t(t-\tau)^{\alpha_{n}-1}f(\tau,u(\tau))d\tau\\
&\leq\sum_{j=1}^{n-1}\frac{M_j B(1-\alpha_j,\alpha_n-1)}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}b\int_0^t(t-\tau)^{\alpha_{n}-\alpha_{j}-1}d\tau\\
&\quad + \frac{\Gamma{(\alpha_n+1)}(1-P_1)b}{\Gamma({\alpha_{n}})}
 \int_0^t(t-\tau)^{\alpha_{n}-1}d\tau\\
&=\sum_{j=1}^{n-1}\frac{M_j B(1-\alpha_j,\alpha_n-1)}{(\alpha_n-\alpha_j)
 \Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}b t^{\alpha_n-\alpha_j}
 +(1-P_1)b t^{\alpha_n}\\
&\leq P_1b+(1-P_1)b=b.
\end{align*}
Thus, $\|Tu\|\leq b$, that is, $T: \overline{K}_b \to \overline{K}_b$. 
Also, $T$ is a completely continuous operator via Lemma \ref{lem4.3}.

Step 2: $\{u\in K(\omega,a,b):\omega(u)>a\} \neq \emptyset$ and
 $\omega(Tu)>a$ for all $u\in K(\omega,a,b)$.
Take $u_0(t)=(a+b)/2$, then $\omega(u_0)=\min_{t \in [l_1,\,l_2]}u_0(t)=(a+b)/2>a$ 
and $\|u_0\|=(a+b)/2<b$. Thus, 
$u_0(t)\in \{u\in K(\omega,a,b):\omega(u)>a\} \neq \emptyset$. 
For each $u\in K(\omega,a,b)$, applying condition $(H_3)$, the definition of
 $K$ and Lemma \ref{lem2.5}, we can show
\begin{align*}
\omega(T u)&= \min_{t \in [l_1,\,l_2]}
\Big(\sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_0^t(t-\tau)^{\alpha_{n}-\alpha_{j}-1}
 u(\tau)h_j(t,\tau)d\tau \\
&\quad + \frac{1}{\Gamma({\alpha_{n}})}\int_0^t(t-\tau)^{\alpha_{n}-1}
 f(\tau,u(\tau))d\tau\Big)\\
&\geq \min_{t \in [l_1,\,l_2]}
\Big(\sum_{j=1}^{n-1}\frac{1}{\Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}
 \int_0^t(t-\tau)^{\alpha_{n}-\alpha_{j}-1}u(\tau)h_j(t,\tau)d\tau\Big)\\
&\quad 
+\min_{t \in [l_1,\,l_2]} \Big(\frac{1}{\Gamma({\alpha_{n}})}
 \int_0^t(t-\tau)^{\alpha_{n}-1}f(\tau,u(\tau))d\tau\Big)\\
&\geq \min_{t \in [l_1,\,l_2]}\Big(\sum_{j=1}^{n-1}
 \frac{m_j B(1-\alpha_j,\alpha_n-1)}{\Gamma({\alpha_{n}})
 \Gamma({1-\alpha_{j}})}\int_0^{l_1}(t-\tau)^{\alpha_{n}-\alpha_{j}-1}
 u(\tau)d\tau\Big)\\
&\quad + \frac{1}{\Gamma({\alpha_{n}})}\int_0^{l_1}(l_1-\tau)^{\alpha_{n}-1}
 f(\tau,u(\tau))d\tau\\
&\geq \min_{t \in [l_1,\,l_2]}\Big(\sum_{j=1}^{n-1}
 \frac{m_j B(1-\alpha_j,\alpha_n-1)}{(\alpha_{n}-\alpha_{j})
 \Gamma({\alpha_{n}})\Gamma({1-\alpha_{j}})}La
 [t^{\alpha_n-\alpha_j}-(t-l_1)^{\alpha_n-\alpha_j}]\Big)\\
&\quad + \frac{1}{\Gamma({\alpha_{n}})}\int_0^{l_1}(l_1-\tau)^{\alpha_{n}-1}
 C_2ad\tau\\
&\geq C_1P_2La+\frac{C_2a}{\Gamma({\alpha_{n}+1})}l_1^{\alpha_{n}}\\
&>C_1P_2La  +{(1-C_1P_2L)a}=a,
\end{align*}
which implies  $\omega(T u)>a$.

Step 3: $\|Tx\|<d$ for all $u\in \overline{K}_d$. Proceeding as step 1,
 we can obtain the result easily by making use of the condition $(H_4)$ 
and Lemma \ref{lem2.5}.

Then we obtain the conclusion of the theorem by employing Theorem \ref{thm3}.
\end{proof}


\begin{corollary}\label{cor4.1}
If the conditions {\rm (H2)} and {\rm (H3)} in Theorem \ref{thm4.1} 
are replaced by
\begin{itemize}
\item[(H2')] $P_1<1$ and
 $ \limsup_{u \to +\infty}\max_{t\in[0,1]}\frac{f(t,u)}{u}<\Gamma{(\alpha_n+1)}(1-P_1)$;

\item[(H3')] $C_1P_2L<1$, and $f(t,u)\geq C_2 a$, for all
 $(t,u)\in[0,1]\times[La,+\infty)$.
\end{itemize}
Then the conclusion of Theorem \ref{thm4.1} holds.
\end{corollary}

\begin{proof}
Since (H2') holds, there exists $0<\sigma<\Gamma{(\alpha_n+1)}(1-P_1)$ and $r_1>0$, 
such that $f(t,u) \leq \sigma u$, for all $u \geq r_1$. Let 
$\beta=\max_{0\leq t \leq r_1}u(t)$, then
$$
0\leq f(t,u) \leq \sigma u+\beta,\quad  0\leq u<+\infty.
$$
Let $b>\max\{{\beta}/\big(\Gamma{(\alpha_n+1)}(1-P_1)-\sigma\big),\ a\}$.
Combining this with condition (H3'), we obtain the conditions
 (H2) and (H3). Therefore, the conclusion of Theorem \ref{thm4.1} holds.
\end{proof}


\begin{corollary}\label{cor4.2}
If the condition {\rm (H4)} in  Theorem \ref{thm4.1} is replaced by
\begin{itemize}
\item[(H4')] $P_1<1$ and 
$ \limsup_{u \to 0^+}\max_{t\in[0,1]}\frac{f(t,u)}{u}<\Gamma{(\alpha_n+1)}(1-P_1)$.
\end{itemize}
Then the conclusion of Theorem \ref{thm4.1} holds.
\end{corollary}



\section{Examples}

To illustrate our main results, we present an example:
\begin{equation} \label{e4.1}
\begin{gathered}
 D_{0+}^{1.5}u(t)-a_3(t)D_{0+}^{0.3}u(t)-a_2(t) D_{0+}^{0.2}u(t)-a_1(t)D_{0+}^{0.1}u(t)
 =f(t,u(t)),\\
 u(0)=u'(0)=0, \quad  0\leq t\leq1,
\end{gathered}
\end{equation}
where
\begin{gather*}
a_1(t)=\ln (t^2+{4}),\quad 
a_2(t)=\frac{1}{8}(\sin t+1), \quad
a_3(t)=\frac{1}{12}\big(\frac{t^2}{t^2+1}+1\big),\\
f(t,u)=(t^2+1)u+\beta, \quad \beta >0.
\end{gather*}
Note that
$$
0\leq {b_1}'(t) <\frac{1}{2\ln{4}}<0.5,\quad
0\leq {b_2}'(t) <0.5,\quad 
0\leq {b_3}'(t)\leq \frac{1}{6}<0.5.
$$
A simple computation shows that $P_1\approx 0.5146$.
 Let $l_1=0.98$, $L=0.95$, $d=2\beta$, $a=20\beta/9$,
$b=4\beta$, $C_1<1$, and $C_2=1.4$. It is easy to check that all the hypotheses
 in Theorem \ref{thm4.1} are satisfied. Thus, 
we conclude that problem \eqref{e4.1} has at least three positive solutions.


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\end{document}